Meromorphic function

A Meromorphic Function on an open subset of $ℂ$ is, in the field of Complex Analysis, a function that is holomorphic except at Isolated singularity. These poles must be isolated. The set of all meromorphic functions on a subdomain $U$ of the complex plane has an advantage over the set of holomorphic functions: it forms not only a ring but also a field. In fact, it can be shown that it is the field of fractions of the ring of holomorphic functions.

Definition

Let $U\subseteq ℂ$ be open. A meromorphic function $f$ on $U$ is a function with a discrete set of singularities $S\subseteq U$ with

1. $f:U\setminus S\to ℂ$ is holomorphic.
2. $f$ has a pole at each point $s\in S$.

We say that $f$ is meromorphic on $U$ and write $f\in ℳ(U)$.

Note that a meromorphic function on $U$ is not defined on the entire set $U$, but only on the complement of a discrete subset.

Remark

In the definition of singularities, the following three types of singularities were mentioned:

• Removable singularities,
• Poles (of order $m$),
• Essential singularities.

Meromorphic functions may only have poles in the set of singularities $S$; they must not have essential singularities.

Properties

1. The sum, difference, and product of two functions $f,g\in ℳ(U)$ are again meromorphic, so $ℳ(U)$ is an algebra over the ring of holomorphic functions on $U$.Let $S$ be the set of poles of $f$ and $T$ the set of poles of $g$. Then $S\cup T$ is a discrete subset of $U$, and $f+g,f-g,fg$ are holomorphic on $U\setminus (S\cup T)$, with removable singularities or poles on $S\cup T$.
2. If $U$ is connected, $f,g\in ℳ(U)$, and $g\ne 0$, then $f/g$ is meromorphic on $U$. In this case, $ℳ(U)$ is a field.Let $S$ be the set of poles of $f$ and $T$ the set of poles of $g$. Since $U\setminus T$ is a domain, the zero set $N$ of $g$ is a discrete subset of $U$, by the Identity Theorem. Now $f/g:U\setminus (S\cup T\cup N)\to ℂ$ is holomorphic, on $S\cup T\cup N$ has $f/g$ removable singularities or poles.
3. Locally, every meromorphic function is a quotient of two holomorphic functions. That is, if $f\in ℳ(U)$ and $z_0\in U$, there exist a neighborhood $V$ of $z_0$ and holomorphic functions $p,q:V\to ℂ$ such that $f=p/q$.A deeper result shows that for domains $U$, such a representation is globally always possible. In this case, the field of meromorphic functions is the field of fractions of the ring of holomorphic functions on $U$.

Equivalent Description as Holomorphic Functions with Values in $\widehat{ℂ}$

Another way to describe meromorphic functions on an open set $U\subseteq ℂ$ is to define them as holomorphic functions with values in the Riemann sphere.

Definition

Let $U\subseteq ℂ$. A function $f:U\to \widehat{ℂ}$ is called "holomorphic" at a point $z_0\in U$ with $f(z_0)\ne \mathrm{\infty }$ if there exists a neighborhood $V$ of $z_0$ such that $f(V)\subseteq ℂ$ and $f{|}_V:V\to ℂ$ is holomorphic at $z_0$.$f$ is called "holomorphic" at a point $z_0$ with $f(z_0)=\mathrm{\infty }$ if $\frac{1}{f}$ is holomorphic at $z_0$ in the above sense.

Poles and Points at Infinity

Let $f:U\to \widehat{ℂ}$ be holomorphic with $f(z_0)=\mathrm{\infty }$. If $f$ is not constant on any neighborhood of $z_0$, then by the Identity Theorem, there exists a neighborhood $V$ of $z_0$ such that $f{|}_{V\setminus z_0}:V\setminus z_0\to ℂ$. Since $f$ is holomorphic at $z_0$, $\frac{1}{f}$ has a power series expansion at $z_0$, say

$\frac{1}{f(z)}=\sum _{k=0}^{\mathrm{\infty }}{a}_k(z-z_0{)}^k$

Let $m:=\min \{k:a_k\ne 0\}$. Then

$\frac{1}{f(z)}=(z-z_0{)}^m\underset{g(z):=}{\underbrace{\sum _{k=0}^{\mathrm{\infty }}{a}_{k+m}(z-z_0{)}^k}}$

Here,$g$ is holomorphic with $g(z_0)\ne 0$, so $1/g$ is holomorphic at $z_0$. It follows that

$f(z)(z-z_0{)}^m=\frac{1}{g(z)},z\in V\setminus \{z_0\}$

thus, $f$ in $z_0$ has a pole of order $m$.

Characterization of Meromorphic Functions

Since poles can be described as points at infinity, we have: A meromorphic function $f$ on $U\subseteq ℂ$ is a holomorphic function $f:U\to \widehat{ℂ}$ whose points at infinity do not accumulate (equivalently, $f$ is not constantly equal to $\mathrm{\infty }$ on any component of $U$).

See also

• Complex Analysis

Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:

• Source: Meromorphe Funktion - URL:

https://de.wikiversity.org/wiki/Meromorphe_Funktion

• Date: 01/07/2024

de:Meromorphe Funktion